Question

question 4
for the following exercise, use a graphing utility to estimate the local extrema of the function and to estimate the intervals on which the function is increasing and decreasing.
$f(t) = 3t^{\frac{1}{3}} - t$
furthest left local extrema: ( , )
furthest right local extrema: ( , )
increasing on the interval:
decreasing on the interval:
you can use your own calculator or the one at the end of the exam.

Explanation:

Step1: Find derivative of $f(t)$

$f'(t) = 3 \cdot \frac{1}{3}t^{-\frac{2}{3}} - 1 = t^{-\frac{2}{3}} - 1 = \frac{1}{t^{\frac{2}{3}}} - 1$

Step2: Find critical points

Set $f'(t)=0$:
$\frac{1}{t^{\frac{2}{3}}} - 1 = 0$
$\frac{1}{t^{\frac{2}{3}}} = 1$
$t^{\frac{2}{3}} = 1$
$t^2 = 1$
$t = 1$ or $t = -1$
Note $f'(t)$ is undefined at $t=0$, but $f(t)$ is defined here; test sign around all critical points.

Step3: Test intervals for $f'(t)$

• For $t < -1$: Let $t=-8$, $f'(-8)=\frac{1}{(-8)^{\frac{2}{3}}}-1=\frac{1}{4}-1=-\frac{3}{4}<0$ (function decreasing)
• For $-1 < t < 0$: Let $t=-0.125$, $f'(-0.125)=\frac{1}{(-0.125)^{\frac{2}{3}}}-1=\frac{1}{0.25}-1=3>0$ (function increasing)
• For $0 < t < 1$: Let $t=0.125$, $f'(0.125)=\frac{1}{(0.125)^{\frac{2}{3}}}-1=\frac{1}{0.25}-1=3>0$ (function increasing)
• For $t > 1$: Let $t=8$, $f'(8)=\frac{1}{8^{\frac{2}{3}}}-1=\frac{1}{4}-1=-\frac{3}{4}<0$ (function decreasing)

Step4: Find local extrema values

• At $t=-1$: $f(-1)=3(-1)^{\frac{1}{3}} - (-1)=3(-1)+1=-2$ (local minimum, furthest left)
• At $t=1$: $f(1)=3(1)^{\frac{1}{3}} - 1=3-1=2$ (local maximum, furthest right)

Answer:

Furthest left local extrema: $(-1, -2)$
Furthest right local extrema: $(1, 2)$
Increasing on the interval: $(-1, 1)$
Decreasing on the interval: $(-\infty, -1) \cup (1, \infty)$